Timeline for Electromagnetic duality and vector/pseudovector transformation properties
Current License: CC BY-SA 4.0
13 events
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| Feb 19 at 15:58 | comment | added | daysofsnow | Having thought about this some more, the conclusion I've arrived at is simply that the duality transformation is a mapping strictly of values. Hence, the second variant in my original question is invalid, because it also implies a mapping of vectorial $\leftrightarrow$ pseudovectorial properties. The easiest way to convince oneself of this is just to do a few drawings of a specific example, keeping in mind that if one starts with (say) an E-field, one necessarily also ends with something that transforms like a vector; even if the values are those of some other H-field (say). | |
| Feb 14 at 4:39 | history | edited | Qmechanic♦ |
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| Feb 14 at 1:34 | answer | added | Nullius in Verba | timeline score: 0 | |
| Feb 13 at 20:28 | comment | added | Cryo | Correct transformations under parity are more fundamental | |
| Feb 13 at 20:27 | comment | added | Cryo | In that sense there is nothing to prove. Either impedance picks up the slack when you use duality, or it's a limitation of duality transformation | |
| Feb 13 at 20:24 | comment | added | Cryo | Ok rotations sorted. Regarding duality, this is simply a mathematical trick, right? Whereas transformations under parity is a necessary condition for Maxwell equations to be tensorial, so I would expect that your impedance will need to transform under inversions in a way that maintains correct transformations of electrical and magnetic fields. Since latter is experimentally falsifiable observation, which was not falsfied | |
| Feb 13 at 9:01 | comment | added | daysofsnow | On re-reading my question, I see that the original wording of "[...] rotation $R$ [...]" could lead to confusion, suggesting that I only cared about proper rotations: I've edited the question to clarify that I intend any affine transformation. | |
| Feb 13 at 9:01 | history | edited | daysofsnow | CC BY-SA 4.0 |
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| Feb 13 at 8:54 | comment | added | daysofsnow | @Cryo: Of course, the question is only relevant for improper rotations, agree. The more specific context for my question is trying to prove the invariance of symmetry eigenvalues $x(g) = \langle \mathbf{B}|g\mathbf{H}\rangle = \langle\mathbf{D}|g\mathbf{E}\rangle$ under duality transformations, i.e., that $x(g) = x'(g)$ (again, for improper $g$; for proper $g$, it follows readily). | |
| Feb 12 at 23:09 | comment | added | Cryo | Finally, I have not used duality much in this context, but I would note that not every single-dimensional number need be a scalar. Charge density is a scalar number, yet you need to transform it carefully when you change coordinates. In order to respect parity transformations of electromagnetic fields, you may well find that impedance may also need to transform like a relative scalar | |
| Feb 12 at 23:02 | comment | added | Cryo | Regarding your transformation question. I would note that rotations are orthogonal transformations, so $det\left(\mathbf{R}\right)=1$. So I would expect electric and magnetic fields to transform the same way under rotations. Things change when you start looking at inversions/reflections that have determinant of -1 | |
| Feb 12 at 23:00 | comment | added | Cryo | You are asking several questions, IMHO. Why are electric and magnetic fields transforming differently? I would probably derive this from showing that four-current (charge and current density) transform as a four-vector, at least of a point charge, which, through Maxwells Equations, forces electric and magnetic fields transform the way the do. | |
| Feb 12 at 13:55 | history | asked | daysofsnow | CC BY-SA 4.0 |